Torque and Rotational Equilibrium

Case III:

(m1)(x1) = (m2)(x2)

Mass D:
(m1)(.33) = (200)(.45)
m1=272.727g
Actual mass: 271.02g

Mass B:
(m1)(.33) = (200)(.42)
m1 = 254.54g
Actual mass: 257.51

Mass R:
(m1)(.33) = (200)(.235)
m1 = 142.42g
Actual mass: 132.40g

Mass E:
(m1)(.33) = (200)(.37)
m1 = 224.24g
Actual mass: 206.14g

Mass F:
(m1)(.33) = (200)(.325)
m1 = 196.97
Actual mass: 196.7

Percent Error:

Mass D:
(272.727-271.02)/(271.02) x100 = .63%

Mass B:
(254.54-257.71)/(257.71) x100 = 1.23%

Mass R:
(142.42-132.40)/(132.40) x100 = 7.57%

Mass E:
(224.24-206.14)/(206.14) x100 = 8.78%

Mass F:
(196.97-196.7)/(196.7) x100 = .137%

Case IV

(m1)(x1) = (m2)(x2)
(100)(.239) = (m2)(.261)
m2 = 91.57g
Mass of meter stick is equal to 96.43g
Percent Error: .05%

Questions

1) A large force can produce little or no torque while a small force can produce a large torque. This is because Torque = Lever arm * Force. Therefore, if there is a large force, to make it a small torque, the lever arm is decreased. For a torque to be large from a small force, the lever arm has to be greater.

2) The downward and upward forces acting on the meter stick cancel each other out, to balance the masses. The forces directly balance each other out, so force of the meter stick is irrelevant. Any change in mass at the fulcrum would not change the balance of the masses.

3) (m1)(x1) = (m2)(x2) + (m3)(x3)
(220)(.35) = (m2)(100) + (.5)(120)
Mass of the meter stick = 170g